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Tag 0AVF

Chapter 15: More on Algebra > Section 15.52: Injective modules

Lemma 15.52.4. Let $R$ be a ring. Let $J$ be an $R$-module. The following are equivalent

  1. $J$ is injective,
  2. $\mathop{\rm Ext}\nolimits^1_R(R/I, J) = 0$ for every ideal $I \subset R$, and
  3. for an ideal $I \subset R$ and module map $I \to J$ there exists an extension $R \to J$.

Proof. We have seen the implication (1) $\Leftrightarrow$ (2) in Lemma 15.52.3. In this proof we will show that (1) $\Leftrightarrow$ (3) which is known as Baer's criterion.

Assume (1). Given a module map $I \to J$ as in (3) we find the extension $R \to J$ because the map $\mathop{\rm Hom}\nolimits_R(R, J) \to \mathop{\rm Hom}\nolimits_R(I, J)$ is surjective by definition.

Assume (3). Let $M \subset N$ be an inclusion of $R$-modules. Let $\varphi : M \to J$ be a homomorphism. We will show that $\varphi$ extends to $N$ which finishes the proof of the lemma. Consider the set of homomorphisms $\varphi' : M' \to J$ with $M \subset M' \subset N$ and $\varphi'|_M = \varphi$. Define $(M', \varphi') \geq (M'', \varphi'')$ if and only if $M' \supset M''$ and $\varphi'|_{M''} = \varphi''$. If $(M_i, \varphi_i)_{i \in I}$ is a totally ordered collection of such pairs, then we obtain a map $\bigcup_{i \in I} M_i \to J$ defined by $a \in M_i$ maps to $\varphi_i(a)$. Thus Zorn's lemma applies. To conclude we have to show that if the pair $(M', \varphi')$ is maximal then $M' = N$. In other words, it suffices to show, given any subgroup $M \subset N$, $M \not = N$ and any $\varphi : M \to J$, then we can find $\varphi' : M' \to J$ with $M \subset M' \subset N$ such that (a) the inclusion $M \subset M'$ is strict, and (b) the morphism $\varphi'$ extends $\varphi$.

To prove this, pick $x \in N$, $x \not \in M$. Let $I = \{f \in R \mid fx \in M\}$. This is an ideal of $R$. Define a homomorphism $\psi : I \to J$ by $f \mapsto \varphi(fx)$. Extend to a map $\tilde\psi : R \to J$ which is possible by assumption (3). By our choice of $I$ the kernel of $M \oplus R \to J$, $(y, f) \mapsto y - \tilde\psi(f)$ contains the kernel of the map $M \oplus R \to N$, $(y, f) \mapsto y + fx$. Hence this homomorphism factors through the image $M' = M + Rx$ and this extends the given homomorphism as desired. $\square$

    The code snippet corresponding to this tag is a part of the file more-algebra.tex and is located in lines 12115–12125 (see updates for more information).

    \begin{lemma}
    \label{lemma-characterize-injective-bis}
    Let $R$ be a ring. Let $J$ be an $R$-module.
    The following are equivalent
    \begin{enumerate}
    \item $J$ is injective,
    \item $\Ext^1_R(R/I, J) = 0$ for every ideal $I \subset R$, and
    \item for an ideal $I \subset R$ and module map $I \to J$
    there exists an extension $R \to J$.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    We have seen the implication (1) $\Leftrightarrow$ (2) in
    Lemma \ref{lemma-characterize-injective}. In this proof
    we will show that (1) $\Leftrightarrow$ (3) which is known
    as Baer's criterion.
    
    \medskip\noindent
    Assume (1). Given a module map $I \to J$ as in (3) we find
    the extension $R \to J$ because the map
    $\Hom_R(R, J) \to \Hom_R(I, J)$ is surjective
    by definition.
    
    \medskip\noindent
    Assume (3). Let $M \subset N$ be an inclusion of $R$-modules.
    Let $\varphi : M \to J$ be a homomorphism. We will show that $\varphi$
    extends to $N$ which finishes the proof of the lemma.
    Consider the set of homomorphisms $\varphi' : M' \to J$
    with $M \subset M' \subset N$ and $\varphi'|_M = \varphi$.
    Define $(M', \varphi') \geq (M'', \varphi'')$ if
    and only if $M' \supset M''$ and $\varphi'|_{M''} = \varphi''$.
    If $(M_i, \varphi_i)_{i \in I}$ is a totally
    ordered collection of such pairs, then we obtain a map
    $\bigcup_{i \in I} M_i \to J$ defined by $a \in M_i$
    maps to $\varphi_i(a)$. Thus Zorn's lemma applies.
    To conclude we have to show that if the pair
    $(M', \varphi')$ is maximal then $M' = N$.
    In other words, it suffices to show, given
    any subgroup $M \subset N$, $M \not = N$ and
    any $\varphi : M \to J$, then we can find
    $\varphi' : M' \to J$ with $M \subset M' \subset N$
    such that (a) the inclusion $M \subset M'$ is strict, and
    (b) the morphism $\varphi'$ extends $\varphi$.
    
    \medskip\noindent
    To prove this, pick $x \in N$, $x \not \in M$.
    Let $I = \{f \in R \mid fx \in M\}$. This is an ideal of $R$.
    Define a homomorphism $\psi : I \to J$ by $f \mapsto \varphi(fx)$.
    Extend to a map $\tilde\psi : R \to J$ which is possible by assumption (3).
    By our choice of $I$ the kernel of
    $M \oplus R \to J$, $(y, f) \mapsto y - \tilde\psi(f)$
    contains the kernel of the map $M \oplus R \to N$,
    $(y, f) \mapsto y + fx$. Hence this homomorphism factors
    through the image $M' = M + Rx$ and this extends the given homomorphism
    as desired.
    \end{proof}

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